1、 SOLUTIONS MANUAL to accompany Digital Signal Processing: A Computer-Based Approach Third Edition Sanjit K. Mitra Prepared by Chowdary Adsumilli, John Berger, Marco Carli, Hsin-Han Ho, Rajeev Gandhi, Chin Kaye Koh, Luca Lucchese, and Mylene Queiroz de Farias Not for sale. 1Chapter 2 2.1 (a) , 81 . 4
2、 , 1396 . 9 , 85 . 22 1 2 1 1 1 = = = x x x (b) . 48 . 3 , 1944 . 7 , 68 . 18 2 2 2 1 2 = = = x x x 2.2 Hence, Thus, = . 0 , 0 , 0 , 1 n n n = . 0 , 0 , 0 , 1 1 n n n . 1 + = n n n x 2.3 (a) Consider the sequence defined by If n 0, then k = 0 is not included in the sum and hence, xn = 0 for n 0. On
3、the other hand, for k = 0 is included in the sum, and as a result, xn =1 for Therefore, . = = n k k n x , 0 n . 0 n . , 0 , 0 , 0 , 1 n n n k n x n k = = = = (b) Since it follows that Hence, = , 0 , 0 , 0 , 1 n n n = . 1 , 0 , 1 , 1 1 n n n . , 0 , 0 , 0 , 1 1 n n n n n = = = 2.4 Recall . 1 n n n =
4、Hence, 3 4 2 2 1 3 + + = n n n n n x ) 4 3 ( 4 ) 3 2 ( 2 ) 2 1 ( 3 ) 1 ( + + = n n n n n n n n . 4 4 3 6 2 5 1 2 + + = n n n n n 2.5 (a) , 4 5 1 2 3 0 2 2 = + = n x n c (b) , 0 0 6 3 1 0 8 7 2 3 = = n y n d (c) , 0 0 3 2 2 1 0 2 5 = = n w n e (d) , 2 7 8 0 1 3 3 2 1 5 4 2 = + = n y n x n u (e) , 0 4
5、 0 3 4 2 15 0 4 = + = n w n x n v (f) , 2 2 10 0 0 5 4 3 4 = + = n w n y n s (g) . 7 5 . 24 8 . 2 0 5 . 3 5 . 10 21 5 . 3 = = n y n r 2.6 (a) , 3 2 1 3 2 1 2 5 3 4 + + + + + + = n n n n n n n x , 5 2 4 7 3 8 1 3 1 6 + + + = n n n n n n n y , 8 5 7 2 5 4 2 3 2 2 3 + + + = n n n n n n n w (b) Recall .
6、 1 = n n n Hence, ) 1 ( ) 1 2 ( 5 ) 2 3 ( 4 n n n n n n n x + + + + + + + = ) 4 3 ( 2 ) 2 1 ( 3 ) 1 ( 2 + n n n n n n Not for sale. 2, 4 2 3 2 2 3 1 3 1 4 2 9 3 4 + + + + + + = n n n n n n n n 2.7 (a) z 1 _ z 1 _ + + h0 h1 h2 xn yn xn-1 xn-2From the above figure it follows that . 2 2 1 1 0 + + = n x
7、 h n x h n x h n y (b) h0 z 1 _ z 1 _ + + z 1 _ z 1 _ + + 11 12 22 21 xn yn xn 1 _ xn 2 _ wn 1 _ wn 2 _ wnFrom the above figure we get ) 2 1 ( 0 21 11 + + = n x n x n x h n w and . 2 1 22 12 + + = n w n w n w n y Making use of the first equation in the second we arrive at ) 2 1 ( 0 21 11 + + = n x n
8、 x n x h n y ) 3 2 1 ( 0 21 11 12 + + + n x n x n x h) 4 3 2 ( 0 21 11 22 + + + n x n x n x h 2 ) ( 1 ) ( 0 22 11 12 21 12 11 ( + + + + + = n x n x n x h. 4 3 ) ( ) 21 22 11 22 21 12 + + + n x n x(c) Figure P2.1(c) is a cascade of a first-order section and a second-order section. The input-output re
9、lation remains unchanged if the ordering of the two sections is interchanged as shown below. z 1 _ z 1 _ + + + + 0.6 0.3 0.2 _ 0.8 0.5 _ yn wn 1 _ wn 2 _ wn z 1 _ + 0.4 xn un yn+1Not for sale. 3 The second-order section can be redrawn as shown below without changing its input- output relation. z 1 _
10、 z 1 _ + + + + 0.6 0.3 0.2 _ 0.8 0.5 _ wn 1 _ wn 2 _ wn xn yn z 1 _ + 0.4 un yn+1 z 1 _ z 1 _The second-order section can be seen to be cascade of two sections. Interchanging their ordering we finally arrive at the structure shown below: z 1 _ z 1 _ + + + + 0.6 0.3 0.2 _ 0.8 0.5 _ xn yn z 1 _ + 0.4
11、un yn+1 z 1 _ z 1 _ xn 1 _ xn 2 _ un 1 _ un 2 _ snAnalyzing the above structure we arrive at , 2 2 . 0 1 3 . 0 6 . 0 + + = n x n x n x n s , 2 5 . 0 1 8 . 0 = n u n u n s n u . 4 . 0 1 n y n u n y + = + From Substituting this in the second equation we get after some algebra . 4 . 0 1 n y n y n u + =
12、 . 2 8 . 0 1 18 . 0 4 . 0 1 + = + n y n y n y n s n y Making use of the first equation in this equation we finally arrive at the desired input-output relation . 3 2 . 0 2 3 . 0 1 6 . 0 3 2 . 0 2 18 . 0 1 4 . 0 + + = + + n x n x n x n y n y n y n y (d) Figure P2.19(d) is a parallel connection of a fi
13、rst-order section and a second-order section. The second-order section can be redrawn as a cascade of two sections as indicated below: z 1 _ z 1 _ + + + 0.3 0.2 _ 0.8 0.5 _ wn 1 _ wn 2 _ wn xn z 1 _ z 1 _ y n 2Not for sale. 4 Interchanging the order of the two sections we arrive at an equivalent str
14、ucture shown below: qn z 1 _ z 1 _ + + _ 0.8 0.5 _ xn + 0.3 0.2 z 1 _ z 1 _ y n 2 y n 1 _ 2 _ y n 2 2Analyzing the above structure we get , 2 2 . 0 1 3 . 0 + = n x n x n q . 2 5 . 0 1 8 . 0 2 2 2 = n y n y n q n y Substituting the first equation in the second we have . 2 2 . 0 1 3 . 0 2 5 . 0 1 8 .
15、0 2 2 2 + = + + n x n x n y n y n y (2-1) Analyzing the first-order section of Figure P2.1(d) given below 0.6 xn z 1 _ + 0.4 un y n 1 un 1 _we get , 1 4 . 0 + = n u n x n u . 1 6 . 0 1 = n u n y Solving the above two equations we have . 1 6 . 0 1 4 . 0 1 1 = n x n y n y (2-2) The output of the struc
16、ture of Figure P2.19(d) is given by n y. 2 1 n y n y n y + = (2-3) From Eq. (2-2) we get 2 48 . 0 2 32 . 0 1 8 . 0 1 1 = n x n y n y and . 3 3 . 0 3 2 . 0 2 5 . 0 1 1 = n x n y n y Adding the last two equations to Eq. (2-2) we arrive at 3 2 . 0 2 18 . 0 1 4 . 0 1 1 1 1 + + n y n y n y n y . 3 3 . 0
17、2 48 . 0 1 6 . 0 + + = n x n x n x (2-4) Similarly, from Eq. (2-1) we get . 3 08 . 0 2 12 . 0 3 2 . 0 2 32 . 0 1 4 . 0 2 2 2 = n x n x n y n y n y Adding this equation to Eq. (2-1) we arrive at 3 2 . 0 2 18 . 0 1 4 . 0 2 2 2 2 + + n y n y n y n y . 3 08 . 0 2 08 . 0 1 3 . 0 + = n x n x n x (2-5) Add
18、ing Eqs. (2-4) and (2-5), and making use of Eq. (2-3) we finally arrive at the input- output relation of Figure P2.1(d) as: . 3 22 . 0 2 56 . 0 1 9 . 0 3 2 . 0 2 18 . 0 1 4 . 0 + + = + + n x n x n x n y n y n y n y Not for sale. 52.8 (a) , 1 3 7 2 3 5 2 4 1 * 1 j j j j j n x + = Therefore . 4 1 5 2
19、2 3 3 7 1 * 1 j j j j j n x + = ( ) , . . . . * , 5 1 5 4 3 5 4 5 1 1 1 2 1 1 j j j j n x n x n x cs + = + = ( ) . . . . . * , 5 2 1 4 5 2 2 4 5 2 5 2 1 1 1 2 1 1 j j j j j n x n x n x ca + + + + = = (b) Hence, and thus, Therefore, . 3 / 2 n j e n x = 3 / * 2 n j e n x = . 2 3 / * 2 n x e n x n j =
20、= ( ) , / * , n x e n x n x n x n j cs 2 3 2 2 2 2 1 2 = = + = and ( ) . * , 0 2 2 2 1 2 = = n x n x n x ca(c) Hence, and thus, Therefore, . 5 / 3 n j e j n x = 5 / * 3 n j e j n x = . 3 5 / * 3 n x e j n x n j = = ( ) , * , 0 3 3 2 1 3 = + = n x n x n x csand ( ) . / * , 5 3 3 3 2 1 3 n j ca e j n
21、x n x n x n x = = = 2.9 (a) . 2 0 3 2 1 5 4 = n x Hence, . 4 5 1 2 3 0 2 = n x Therefore, 2 5 2 4 2 5 2 ) ( 2 1 2 1 = + = n x n x n x ev 1 5 . 2 1 2 1 5 . 2 1 = and 6 5 4 0 4 5 6 ) ( 2 1 2 1 = = n x n x n x od. 3 5 . 2 2 0 2 5 . 2 3 = (b) . 2 7 8 0 1 3 6 0 0 0 0 = n y Hence, . 0 0 0 0 6 3 1 0 8 7 2
22、= n y Therefore, 1 5 . 3 4 0 5 . 2 3 5 . 2 0 4 5 . 3 1 ) ( 2 1 = + = n y n y n y evand . 1 5 . 3 4 0 5 . 3 0 5 . 3 0 4 5 . 3 1 ) ( 2 1 = = n y n y n y od(c) . 5 2 0 1 2 2 3 0 0 0 0 0 0 0 0 0 0 = n w Hence, . 0 0 0 0 0 0 0 0 0 0 3 2 2 1 0 2 5 = n w Therefore ) ( 2 1 n w n w n w ev + = Not for sale. 6
23、 5 . 2 1 0 5 . 0 1 1 5 . 1 0 0 0 5 . 1 1 1 5 . 0 0 1 5 . 2 = and ) ( 2 1 n w n w n w od = . 5 . 2 1 0 5 . 0 1 1 5 . 1 0 0 0 5 . 1 1 1 5 . 0 0 1 5 . 2 = 2.10 (a) Hence, . 2 1 + = n n x . 2 1 + = n n x Therefore, = + + + = , 3 , 2 / 1 , 2 2 , 1 , 3 , 2 / 1 ) 2 2 ( 2 1 , 1 n n n n n n x evand = + + = .
24、 3 , 2 / 1 , 2 2 , 0 , 3 , 2 / 1 ) 2 2 ( 2 1 , 1 n n n n n n x od(b) Hence, Therefore, . 3 2 = n n x n . 3 2 = n n x n () = + = , 3 , , 2 2 , 0 , 3 , 3 3 2 1 2 1 2 1 , 2 n n n n n n x n n n n evand () = = . 3 , , 2 2 , 0 , 3 , 3 3 2 1 2 1 2 1 , 2 n n n n n n x n n n n od(c) Hence, Therefore, . 3 n n
25、 n x n = . 3 n n n x n = () n n n ev n n n n n n x = + = 2 1 ) ( 2 1 , 3and () . ) ( 2 1 2 1 , 3 n n n od n n n n n n x = = (d) . 4 n n x = Hence, . 4 4 n x n x n n = = = Therefore, n ev n x n x n x n x n x n x = = + = + = ) ( ) ( 4 4 4 2 1 4 4 2 1 , 4and . 0 ) ( ) ( 4 4 2 1 4 4 2 1 , 4 = = = n x n
26、x n x n x n x od2.11 () . 2 1 n x n x n x ev + = Thus, () . 2 1 n x n x n x n x ev ev = + = Hence, is an even sequence. Likewise, n x ev () . 2 1 n x n x n x od = Thus, () . 2 1 n x n x n x n x od od = = Hence, is an odd sequence. n x odNot for sale. 72.12 (a) Thus, . n x n x n g ev ev = . n g n x n
27、 x n x n x n g ev ev ev ev = = = Hence, is an even sequence. n g(b) Thus, . n x n x n u od ev = ( ) . n u n x n x n x n x n u od ev od ev = = = Hence, is an odd sequence. n u(c) Thus, . n x n x n v od od = ( )( ) n x n x n x n x n v od od od od = = Hence, is an even sequence. . n v n x n x od od = =
28、 n v2.13 (a) Since is causal, n x . 0 , 0 = n n x Now, () . 2 1 n x n x n x ev + = Hence, () 0 0 0 0 2 1 x x x x ev = + = and . 0 , 2 1 = n n x n x evCombining the two equations we get = . 0 , 0 , 0 , , 0 , 2 n n n x n n x n x ev evLikewise, () . 2 1 n x n x n x od = Hence, () 0 0 0 0 2 1 = = x x x
29、odand . 0 , 2 1 = n n x n x odCombining the two equations we get = . 0 , 0 , 0 , 2 n n n x n x ev(b) Since is causal, n y . 0 , 0 = n n y Let where and are real causal sequences. , n jy n y n y im re + = n y re n y im Now, ( ) . 2 1 n y n y n y ca = Hence, ( ) 0 0 0 0 2 1 im ca jy y y y = = and . 0
30、, 2 1 = n n n y y caSince is not known, cannot be fully recovered from . 0 re y n y n y caLikewise, ( ) . 2 1 n y n y n y cs + = Hence, ( ) 0 0 0 0 2 1 re cs y y y y = + = and . 0 , 2 1 = n n n y y csSince is not known, cannot be fully recovered from . 0 im y n y n y cs2.14 Since is causal, n x . 0
31、, 0 = = 2.15 (a) where n A n x = A and are complex numbers with . 1 Since for n n , 0 can become arbitrarily large, is not a bounded sequence. n xNot for sale. 8(b) where Since for n n , 0 can become arbitrarily large, is not a bounded sequence. n h(d) Since ). cos( 4 n n g o = 4 n g for all values
32、of is a bounded sequence. , n g n(e) = . 0 , 0 , 1 , 1 2 1 n n n v nSince 1 2 1n 1 2 1 = nfor , 1 = n 1 n v for all values of Thus is a bounded sequence. . n n v2.16 . 1 ) 1 ( 1 = + n n n x nNow . 1 ) 1 ( 1 1 1 = = = = = + = n n n n n n n x Hence is not absolutely summable. n x2.17 (a) Now . 1 1 = n
33、 n x n = = = = = = 1 1 1 2 n n n n n n x , since . 1 Hence, is absolutely summable. 1 n x(b) Now . 1 2 = n n x n = = = = = 1 1 2 n n n n n n n n x , ) 1 ( 2 = since . 1 2 Hence, is absolutely summable. 2 n x(c) Now . 1 2 3 = n n n x n n n n n n n n n x = = = = = 1 2 1 2 3 K + + + + = 4 2 3 2 2 2 4 3
34、 2 ) ( 5 ) ( 3 ) ( 5 4 3 4 3 2 4 3 2 K K K + + + + + + + + + + + + = Not for sale. 9) ( 7 6 5 4 K + + + + = K + + + + 1 7 1 5 1 3 1 4 3 2 = = = = = 1 1 1 2 1 1 ) 1 2 ( 1 1 n n n n n n n n = 1 ) 1 ( 2 1 1 2. ) 1 ( ) 1 ( 3 + = Hence, is absolutely summable. 3 n x 2.18 (a) . 2 1 n n x n a = Now . 2 1 1
35、 2 1 2 1 2 1 0 0 = = = = = = = n n n n n a n x Hence, is absolutely summable. n x a(b) . ) 2 )( 1 ( 1 n n n n x b + + = Now + + = = = 0 ) 2 )( 1 ( 1 n n b n n n x . 1 5 1 4 1 4 1 3 1 3 1 2 1 2 1 1 2 1 1 1 0 = + + + + = + + = = K n n nHence, is absolutely summable. n x b2.19 (a) A sequence is absolut
36、ely summable if n x . = n n x By Schwartz inequality we have . 2 = = = n n n n x n x n x Hence, an absolutely summable sequence is square summable and has thus finite energy. Now consider the sequence . 1 1 = n n x nThe convergence of an infinite series can be shown via the integral test. Let ), (x
37、f a n = where a continuous, positive and decreasing function is for all Then the series and the integral both converge or both diverge. For . 1 x =1 n n a 1 ) ( dx x f . ) ( , 1 1 x n n x f a = = But = = = 0 ) (ln 1 1 1 x dx x . Hence, = = = 1 1 n n n n x does not converge. As a result, 1 1 = n n x nis not absolutely summable. Not for sale. 10 (b) To show that is square-summable, we observe that here n x 2
